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Trains
I want to construct a train of total length 12 units using cars
which are either 1 unit long or 2 units long. The question is,
how many different trains are there? For example, here is one
possibility, using 6 cars of length 1 and 3 cars of length 2.
1
1
1
2
2
1
2
1
1
Other possibilities are obtained by rearranging the cars, or using
different numbers of short and long cars. Note that a train has a
front and a back, so that the mirror image of the train diagrammed above
1
1
2
1
2
2
1
1
1
Of course, we are not so
much interested in trains of
length 12 as in trains of any
length n. That is we want to
discover the relationship between the length of the train
and the number of different
trains that can be constructed.
But 12 serves as a good concrete starting point—small
enough that the student will
start to look at possibilities,
but too large to be easily
solved directly.
is counted as a different train.
The students often seem to spend the first ten minutes blasting
away at the problem--organizing the different types of trains by
some scheme, often according to the number of 2-cars, trotting
out the combinatorial coefficients, etc. After a while I say: "what
makes the problem hard? Too many possibilities! How could
we make life simpler? Have a shorter train?
Indeed, they recall what they've been always told––find a simpler problem, and work on it first.
Okay: how many trains of length 1? Clearly 1 such train—a single 1-car. How many trains of length 2? Clearly 2 such trains—
a single 2-car, or two 1-cars. How many trains of length 3?––of
length 4? We have to start making a systematic list of the possibilities. I let them work on that for a moment and collect on the
board the results for trains of length up to 6.
We stare at the results and you can hear the smiles breaking out
all over the classroom.
Indeed we have the Fibonacci numbers!
Indeed there are some interesting direct approaches
which stay with trains of
length 12. But my agenda
here is to get them to work up
from short trains to large.
That’s often an effective
mathematical strategy
For example, here are
the trains of length 5
1-1-1-1-1
1-1-1-2
1-1-2-1
1-2-1-1
2-1-1-1
1-2-2
2-1-2
2-2-1
There are a total of 8 of
them.
trains 1
If we let tn be the number of trains of length n, then the sequence
tn appears to be just the Fibonacci sequence un. Well not
quite—it’s shifted by one—for example, the 5th train number t5
is 8 but that’s the 6th Fibonacci number u6 . But the train numbers do appear to follow the same additive rule as the Fibonacci
sequence:
t n = t n −1 + t n −2 .
Well!––if this holds, we have a nice arithmetic way of calculating t12 ––just iterate the additive rule till we get to n=12. When
we do that, we find that there are 233 trains of length 12.
So are we finished?—how many think that t12 is 233?
My students are wary as usual, and they need to be encouraged a
bit, but eventually most of them vote yes.
Well I agree, I think it is too. BUT—can we be sure? Have we
shown it beyond doubt?
Some care is needed here. The arithmetic pattern we found in
the first six train numbers is certainly compelling, but can we be
sure that it will continue to hold? Well not yet. There are nice
examples of compelling patterns that suddenly fail (see Regions). We really need an argument.
Length
of train
1
2
3
4
5
6
n
1
2
3
4
5
6
7
8
9
10
11
12
Number of
trains
1
2
3
5
8
13
tn
1
2
3
5
8
13
21
34
55
89
144
233
So that's our job: to find some way of convincing ourselves that
this simple sum rule should hold for the train sequence. As an
example, I give the class the following specific task—convince
me that there are 13 = 8+5 trains of length 6 without actually
listing them, but using the fact that there are 8 trains of length 5
and 5 trains of length 4.
Well the argument they come up with, briefly stated, is that you
can make a 6-train by starting with any 5-train and adding a car
of length 1, or by starting with any 4-train and adding a car of
length 2, and there will be 8 trains of the first type and 5 trains of
the second type, for a total of 13 trains.
Right? Well it’s a nice argument, but it requires just a bit more
care. What they’ve given me is two procedures for constructing
a 6-train, one that starts with a 5 train, and the other that starts
with a 4-train. For this to actually give an exact count of the 6trains you really have to check first that you get all of the 6trains in one of these two ways, and secondly, that you haven’t
counted any 6-train twice, that is, that you can’t get the same 6train coming out of both routes.
trains 2
Neither of these are hard to do, but the point is that such a check
needs to be made to be sure that the argument works.
To help them “see” how to do this, I give them the following
challenge. You want to show that
t6 = t5 + t4
That is, you want to show that the number of 6-trains is the sum
of two known numbers. Well to do this what you really want to
do is partition the set of 6-trains into two natural subsets whose
size is those two known numbers. For example imagine you’re
on top of the CN tower looking down at all the 6 trains arrayed
in the railroad yard. You want to find some natural way to sort
them into two subsets, one of which has size t 5 and the other of
which has size t 4 .
Well, here’s how to do it. Let one subset be those trains that begin with a 1-car and let the other subset be those trains that begin
with a 2-car.
Every train is clearly in exactly one of those two sets. Now how
many trains are there in each set? Clearly there are t 5 in the first
(the rest of the train has length 5) and t 4 in the second (the rest
of the train has length 4). So t 6 must be the sum of those two
numbers.
It’s clear that this argument is quite general, and could be used to
show that the number of 7-trains was equal to the sum of the
number of 6-trains and the number of 5-trains etc. So the additive rule always holds, and our conclusion that t 12 =233 is correct.
Now there's something really neat that we can do with this.
What we have obtained is a physical "model" of the Fibonacci
numbers, and the idea is to see if we can use that to obtain proofs
of arithmetic relationships among these numbers.
We have shown that the
train numbers are really
the same as the Fibonacci
numbers. Now there’s an
awesome payoff for this
connection. Read on.
trains 3
A "train-theoretic" proof of a Fibonacci identity.
As an example, take the "sum-of-squares" identity, of which an
example is
u 11 = u 6 2 + u 5 2 .
There are quite a few ways we might try to show that this pattern
holds in general (for example, by induction––see Fibonacci
problem 2), but here’s a new way—don’t think about Fibonacci
numbers at all—instead, think about trains. First we formulate
the identity in “train language”:
t 10 = t 5 2 + t 4 2 .
Now we ask––can we find a "train-theoretic" argument for this
identity?
This time, we want to partition the 10-trains into two disjoint
classes so that there are t 52 trains of the first kind and t 42 trains
of the second kind. A clue comes from noting that the subscripts
5 and 4 have the status of being roughly half of 10––this suggest
that the classification ought to be based on something like "cutting the 10-trains in half." Can you take it from there?
Well, here it is. There are two kinds of 10-trains, depending on
whether or not they can be “cut in half” Those that can must
have a join in the middle, and those that cannot must have a 2car in the middle.
5
4
5
2
This is another good small
group problem for the class,
but this time I find they have
trouble with it. In fact, as I
wander around, I discover a
lot of frustration. They know
what they have to do––look at
half the train—but they just
can’t seem to get it together.
Typically they try to start with
all the 4-trains and all the 5trains and build 10-trains out
of them. “You have to start
with 10-trains,” I tell them,
“and identify two types, and
give me a clear rule by which
I’ll be able to classify a 10train as being of one type or
another.
4
Those with a join are constructed of two trains of length 5, and
there are t5 possibilities for the front part and another t5 possibilities for the back, for a total of t 52 trains with a join in the middle.
Those with a 2-car in the middle are completed by adding two
trains of length 4, and there are t 42 possibilities for that. Adding
these up, we get
t10 = t 5 2 + t 4 2 .
Now that's an argument of great beauty!
trains 4
Problems
1. For the sequence that’s begun at the right, each term is equal
to the previous term plus twice the term before that. If the first
and second terms are both 1, find the 8th term x8.
n
1
2
3
4
xn
1
1
3 =1+2×1
5 =3+2×1
n
1
2
3
4
xn
4
10
6 =10–4
4 =10–6
n
1
2
3
4
xn
1
2
3 =2×2–1
4 =2×3–2
2. Use the same rule as 1) but take x1 = 1 and x2 = 2. What now
is the 8th term?
3. . For the sequence that’s begun at the right, each term is equal
to the magnitude of the difference between the two preceding
terms. If the first term is 4 and the second term is 10, what is the
100th term?
4. For the sequence that’s begun at the right, each term is equal
to twice the previous term minus the term before that. If the first
term is 1 and second term is 2, find the 100th term x100.
5. Suppose you take the rule of #4 but start with a different pair
of numbers. Experiment a bit and see what kind of patterns you
get. Can you formulate a general hypothesis and even perhaps
see why it might hold?
6. Let sn be the number of different symmetric trains of length n
which are built out of cars which are either 1 unit long or 2 units
long––where, a symmetric train is one that is the same running
backwards and forwards. Thus 212 is a symmetric 5-train but
122 is not. For example, s5 = 2 (table at the right). Find a formula for sn in terms of the Fibonacci numbers. Can you find a
general argument for your formula?
7. Let rn be the number of different reversible trains of length n
which are built out of cars which are either 1 unit long or 2 units
long. This is the same as the standard train problem, except the
5-trains 122 and 221 are no longer considered different, because
one is the same as the other in reverse order. But the train 212 is
different from these. For example, r5 = 5 (table at the right).
Find a formula for rn in terms of the Fibonacci numbers. Can
you find a general argument for your formula?
The 2 symmetric trains of
length 5
1-1-1-1-1
2-1-2
The 5 reversible trains of
length 5
1-1-1-1-1
1-1-1-2
1-1-2-1
1-2-2
2-1-2
8. Give a "train-theoretic" proof of the Fibonacci identity:
u11 = u 4 ⋅ u 6 + u 5 ⋅ u 7 .
trains 5
1
9. Consider Pascal's triangle and
observe that the Fibonacci numbers are the sums of the shallow
diagonals (these are in 1-1 correspondance with the left-hand 1's).
The combinatorial formula for this is
un+1
⎛ n⎞ ⎛ n − 1⎞ ⎛ n − 2⎞ ⎛ n − 3⎞
= ⎜ ⎟ +⎜
⎟ +...
⎟ +⎜
⎟ +⎜
⎝ 0⎠ ⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠
1
1
1
1
1
1
1
1
1
1
9
10
5
7
8
20
56
15
70
1
6
21
56
126
252
1
5
35
126
210
1
4
10
35
84
120
6
15
28
1
3
10
21
36
45
3
4
6
1
2
1
7
28
84
210
1
8
36
120
1
9
45
1
10
There is a very elegant "traintheoretic" proof of this formula. Find
it.
10. Give a "train-theoretic" proof of each of the following Fibonacci identities. The identities are
quite general, but I give particular versions. In (b) there is a different form for odd and even n.
u13 = u11 + u10 + u9 +...+u1 + 1.
(a)
u12 = u11 + u9 + u7 +...+u1 . and u13 = u12 + u10 + u8 +...+u2 + 1 .
(b)
11. Give a "train-theoretic" proof of the Fibonacci identity: u11 = u4 ⋅ u6 + u5 ⋅ u7 .
12.(a) Let sn be the number of sequences of 0's and 1's of length n which have no consecutive
1’s. For example, there are a total of 8 sequences of 0’s and 1’s of length 3 (why?), but only 5 of
these have no consecutive 1’s. Thus s3 = 5. Find the sequence sn.
(b) A fair coin is tossed until two consecutive heads appear. This might take 2 tosses, or three, or
four, etc. Let pn be the probability that it will take exactly n tosses. Thus p3 = 1/8 because only
1 of the 8 possible sequences of H and T gets two consecutive heads for the first time exactly on
the 3rd toss (that’s THH). Find the sequence pn.
(c) If you've solved (b) you have a sequence pn of interesting numbers. Now sooner or later you
expect to get two heads in a row, so the sum of all the p n (2 ≤ n ≤ ∞) must equal 1. That gives
an interesting summation formula, and a proof of it at the same time. What is it? Can you find an
alternative (direct) proof?
13. Give a "train-theoretic" proof of each of the following Fibonacci identities. The identities are
quite general, but I give particular versions. In (b) there’s a different form for odd or even n.
2
2
u12 u11 = u11
+ u10
+ u92 +...+u12
(a)
(b)
2
u13
= u13 u12 + u12 u11 + u11u10 +...+u2 u1 + 1
2
u12
= u12 u11 + u11u10 + u101u9 +...+u2 u1
[Hint: These are more complicated than #2 in that they involve working with two trains together.
For example in (a) we want an 11-train and a 10-train. Park the two trains parallel to one another
with the fronts lined up. Now start at the back and walk towards the front between the two trains.
Stop when you come to the first 1-car, on either train, and classify the situation according to the
position of this car. Do the same thing for (b), except that now the trains are the same length, so
that the counting is a bit trickier. It's very satisfying when you see how to make it work.
trains 6
1